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Trigonometrical Equations
hard
The only value of $x$ for which ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$ holds, is
A
$\frac{{5\pi }}{4}$
B
$\frac{{3\pi }}{4}$
C
$\frac{\pi }{2}$
D
All values of $x$
Solution
(a) Since $A.M.$ $ \ge $ $G.M.$
$\Rightarrow$ $\frac{1}{2}({2^{\sin x}} + {2^{\cos x}}) \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $
$ \Rightarrow $ ${2^{\sin x}} + {2^{\cos x}} \ge {2.2^{\frac{{\sin x + \cos x}}{2}}}$
$ \Rightarrow $${2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \frac{{\sin x + \cos x}}{2}}}$
and we know that $\sin x + \cos x \ge – \sqrt 2 $
$\therefore $ ${2^{\sin x}} + {2^{\cos x}} > {2^{1 – (1/\sqrt 2 )}}$, for $x = \frac{{5\pi }}{4}$.
Standard 11
Mathematics
